# How do you solve #log_5(3x-1)=log_5 (2x^2)# and check the solutions?

##### 1 Answer

Dec 18, 2016

#### Explanation:

We start by using the property that if

Therefore:

#3x - 1 = 2x^2#

#0 = 2x^2 - 3x + 1#

#0 = 2x^2 - 2x - x + 1#

#0 = 2x(x- 1) - 1(x - 1)#

#0 = (2x- 1)(x - 1)#

#x = 1/2 and 1#

Checking in the original equation, you will find **both** solutions work. Note that our restrictions on the variable are *equal to or smaller than #0#*. We can automatically confirm both solutions are correct because neither contradict the restriction.

Hopefully this helps!